By Johnny Henderson, Rodica Luca
Boundary worth difficulties for platforms of Differential, distinction and Fractional Equations: confident suggestions discusses the idea that of a differential equation that brings jointly a suite of extra constraints referred to as the boundary conditions.
As boundary price difficulties come up in different branches of math given the truth that any actual differential equation could have them, this ebook will offer a well timed presentation at the subject. difficulties related to the wave equation, akin to the selection of ordinary modes, are frequently said as boundary price difficulties.
To be worthwhile in functions, a boundary worth challenge can be good posed. which means given the enter to the matter there exists a distinct resolution, which relies consistently at the enter. a lot theoretical paintings within the box of partial differential equations is dedicated to proving that boundary worth difficulties bobbing up from medical and engineering purposes are actually well-posed.
- Explains the structures of moment order and better orders differential equations with indispensable and multi-point boundary conditions
- Discusses moment order distinction equations with multi-point boundary conditions
- Introduces Riemann-Liouville fractional differential equations with uncoupled and matched necessary boundary conditions
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Additional info for Boundary value problems for systems of differential, difference and fractional equations : positive solutions
7, for any t ∈ [σ , 1 − σ ] we obtain (Du)(t) ≥ C8 1−σ G1 (t, s) σ 1−σ ≥ C8 ν1 σ β 1−σ σ β1 G2 (s, τ )g(τ , u(τ )) dτ 1−σ J1 (s) (ε1 ν2 )β1 β β ≥ C8 ν1 ν2 1 ε1 1 ν β1 β2 θ1 θ2 1 u σ β1 β2 ds J2 (τ )(u(τ ))β2 dτ β1 ds ≥ u . Therefore, Du ≥ u , ∀ u ∈ ∂Bε2 ∩ P0 . 1 (2), we deduce that D has at least one fixed point u2 ∈ (B¯ R1 \ Bε2 ) ∩ P0 . Then our problem (S )–(BC) has at least 1 one positive solution (u2 , v2 ) ∈ P0 × P0 , where v2 (t) = 0 G2 (t, s)g(s, u2 (s)) ds. 2. 3 Examples In this section, we shall present two examples which illustrate our results above.
By using (H5), we also have v1 > 0. If we suppose that v1 (t) = 0 for all t ∈ [0, 1], then by using (H5), we have f (s, v1 (s)) = f (s, 0) = 0 for all s ∈ [0, 1]. This implies u1 (t) = 0 for all t ∈ [0, 1], which contradicts u1 > 0. 4 is completed. 5. Assume that (H1)–(H5) hold. If the functions f and g also satisfy the following conditions (H8) and (H9), then problem (S )–(BC) has at least one positive solution (u(t), v(t)), t ∈ [0, 1]: Systems of second-order ordinary differential equations 31 (H8) There exist α1 , α2 > 0 with α1 α2 ≤ 1 such that f (t, u) s (1) f˜∞ = lim sup sup ∈ [0, ∞) α u→∞ t∈[0,1] u 1 and (2) g˜ s∞ = lim sup sup u→∞ t∈[0,1] g(t, u) = 0.
58) satisfies αT + β d αξj + β u(ξj ) ≥ d T u(t) ≤ (T − s)y(s) ds, 0 ≤ t ≤ T, (T − s)y(s) ds, ∀ j = 1, . . , m. 5 (Li and Sun, 2006). Assume that α ≥ 0, β ≥ 0, ai ≥ 0 for all m i = 1, . . , m, 0 < ξ1 < · · · < ξm < T, T > i=1 ai ξi > 0, d > 0, and y ∈ 1 C(0, T) ∩ L (0, T), y(t) ≥ 0 for all t ∈ (0, T). 58) satisfies inft∈[ξ1 ,T] u(t) ≥ r1 supt ∈[0,T] u(t ), where r1 = min m ai (T − ξi ) ξ1 , i=1 m , T T − i=1 ai ξi m i=1 ai ξi T , s−1 i=1 ai ξi + T− m i=s ai (T m i=s ai ξi − ξi ) , s = 2, .